Chemistry - Carbonyl compounds Question with Solution | TestHub

Both acetone and acetaldehyde form coloured 2,4-DNP hydrazones. Benzaldehyde lacks $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{O}}^{-}$ and is iodoform-negative; both glucose and fructose give Fehling tests because fructose enediolises in alkali. Strictly, crossed Cannizzaro is oxidation of one aldehyde species and reduction of another, rather than disproportionation of a single species. Thus D is the unambiguous correct statement; some loose textbook terminology may call all Cannizzaro reactions disproportionation.